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Let KKK be a field and EEE an extension field of KKK. If α∈EαE\alpha\in E, then the smallest subfield of EEE, that contains KKK and
αα\alpha, is denoted by K⁢(α)KαK(\alpha). We say that K⁢(α)KαK(\alpha) is
obtained from the field KKK by adjoining the element αα\alpha
to KKK via field adjunction.

Theorem.K⁢(α)KαK(\alpha) is identical with the quotient fieldQQQ of K⁢[α]KαK[\alpha].

Proof. (1) Because K⁢[α]KαK[\alpha] is an integral domain (as a subring of the field EEE), all possible quotients of the elements of K⁢[α]KαK[\alpha] belong to EEE. So we have

and because K⁢(α)KαK(\alpha) was the smallest, then K⁢(α)⊆Q.KαQK(\alpha)\subseteq Q.

(2) K⁢(α)KαK(\alpha) is a subring of EEE containing KKK and αα\alpha, therefore also the whole ring K⁢[α]KαK[\alpha], that is, K⁢[α]⊆K⁢(α)KαKαK[\alpha]\subseteq K(\alpha). And because K⁢(α)KαK(\alpha) is a field, it must contain all possible quotients of the elements of K⁢[α]KαK[\alpha], i.e., Q⊆K⁢(α)QKαQ\subseteq K(\alpha).

In addition to the adjunction of one single element, we can adjoin to KKK an arbitrary subset SSS of EEE: the resulting field K⁢(S)KSK(S) is the smallest of such subfields of EEE, i.e. the intersection of such subfields of EEE, that contain both KKK and SSS. We say that K⁢(S)KSK(S) is obtained from KKK by adjoining the set SSS to it. Naturally,

K⊆K⁢(S)⊆E.KKSEK\subseteq K(S)\subseteq E.

The field K⁢(S)KSK(S) contains all elements of KKK and SSS, and being a field, also all such elements that can be formed via addition, subtraction, multiplication and division from the elements of KKK and SSS. But such elements constitute a field, which therefore must be equal with K⁢(S)KSK(S). Accordingly, we have the

Theorem.K⁢(S)KSK(S) constitutes of all rational expressions formed of the elements of the field KKK with the elements of the set SSS.

Notes.

1. K⁢(S)KSK(S) is the union of all fields K⁢(T)KTK(T) where TTT is a finite subset of SSS.
2. K⁢(S1∪S2)=K⁢(S1)⁢(S2)KsubscriptS1subscriptS2KsubscriptS1subscriptS2K(S_{1}\cup S_{2})=K(S_{1})(S_{2}).
3. If, especially, SSS also is a subfield of EEE, then one may denote K⁢(S)=K⁢SKSKSK(S)=KS.